11
$\begingroup$

How do we evaluate the following series:

$$\sum_{n=1}^\infty \frac1{4n^2+2n}$$

I know that it converges by the comparison test. Wolfram Alpha gives the answer $1 - \ln(2)$, but I cannot see how to get it. The Taylor series of logarithm is nowhere near this series.

$\endgroup$
  • 3
    $\begingroup$ This series is actually one minus the alternating harmonic series, which is known to converge to $\ln(2)$. $\endgroup$ – Mark McClure May 30 '13 at 17:21
  • 2
    $\begingroup$ After all these answers, I'm still the only person who's up-voted the question. $\endgroup$ – Michael Hardy May 30 '13 at 17:53
  • $\begingroup$ Thanks for the nice edits and the up-vote too! $\endgroup$ – Vishal Gupta May 31 '13 at 4:33
  • 2
    $\begingroup$ @Vishal Just a note about mathematical English: The term "series" already incorporates summation. Therefore, one speaks of "evaluating a series" rather than "summing a series". $\endgroup$ – Lord_Farin May 31 '13 at 8:58
11
$\begingroup$

Rewrite the series as follows:

$$\sum_{n=1}^\infty\frac{1}{4n^2+2n}=\sum_{n=1}^\infty\frac{1}{2n(2n+1)}=\sum_{n=1}^\infty\left(\frac{1}{2n}-\frac{1}{2n+1}\right)=\sum_{m=2}^\infty\frac{(-1)^m}{m}.$$

You may now evaluate it using the Taylor series of logarithm.

Added: The third equality is justified as follows:

Write $a_N=\sum_{n=1}^N\left(\dfrac{1}{2n}-\dfrac{1}{2n+1}\right)$ and $b_M=\sum_{m=2}^M\dfrac{(-1)^m}{m}$. The sequence of partial sums $b_M$ is convergent by the Leibniz criterion. Furthermore, $a_N = b_{2N+1}$ holds for all $N\in\mathbb N$, i.e. $(a_N)_{N=1}^\infty$ is a subsequence of $(b_M)_{M=1}^\infty$. Therefore these sequences converge to the same limit, which justifies the equality.

$\endgroup$
  • $\begingroup$ Does it not involve rearranging the series? $\endgroup$ – Vishal Gupta May 31 '13 at 4:56
  • $\begingroup$ @Vishal: Not really. See the justification I added. $\endgroup$ – Dejan Govc May 31 '13 at 8:36
  • $\begingroup$ Thanks, that is a nice trick $\endgroup$ – Vishal Gupta Jun 8 '13 at 3:52
8
$\begingroup$

$$\dfrac1{4n^2+2n} = \dfrac1{2n(2n+1)} = \dfrac1{2n} - \dfrac1{2n+1}$$ Hence, \begin{align} \lim_{m \to \infty} \sum_{n=1}^{m} \dfrac1{4n^2+2n} & = \lim_{m \to \infty} \sum_{n=1}^{m} \left(\dfrac1{2n} - \dfrac1{2n+1} \right)\\ & = \lim_{m \to \infty}\left(\dfrac12 - \dfrac13 + \dfrac14 - \dfrac15 \pm \cdots +\dfrac1{2m} - \dfrac1{2m+1}\right)\\ & = 1 - \lim_{m \to \infty}\left(1-\dfrac12 + \dfrac13 - \dfrac14 + \dfrac15 \mp \cdots -\dfrac1{2m} + \dfrac1{2m+1}\right)\\ & = 1 - \lim_{m \to \infty}\left(1-\dfrac12 + \dfrac13 - \dfrac14 + \dfrac15 \mp \cdots -\dfrac1{2m} \right) - \lim_{m \to \infty}\dfrac1{2m+1}\\ & = 1 - \lim_{m \to \infty}\dfrac1{2m+1} - \lim_{m \to \infty} \sum_{n=1}^{2m} \dfrac{(-1)^{n-1}}{n}\\ & = 1 - 0 - \log(1+1)\,\,\, \left\{\text{Since $\log(1+x) = -\displaystyle \sum_{n=1}^{\infty}\dfrac{(-x)^{n}}n$} \right\}\\ & = 1 - \log(2) \end{align}

$\endgroup$
7
$\begingroup$

Hint: $\frac 1{4n^2+2n} = \frac 1{2n}-\frac 1{2n+1}$

$\endgroup$
6
$\begingroup$

Knowing the answer is helpful. Recall that $$\ln 2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots.$$ Thus $$\ln 2=1-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-\cdots.$$ The "typical" term $\dfrac{1}{2k}-\dfrac{1}{2k+1}$ simplifies to $\dfrac{1}{2k(2k+1)}$.

$\endgroup$
  • $\begingroup$ Are you not rearranging the series? $\endgroup$ – Vishal Gupta May 31 '13 at 4:57
  • $\begingroup$ @Vishal He's actually taking a subsequence of the partial sums. That is OK. $\endgroup$ – Pedro Tamaroff Sep 3 '13 at 22:07
1
$\begingroup$

$\gamma\,$: Euler-Mascheroni constant. $\quad\Psi\left(z\right)\,$: Digamma function.

\begin{align} &\\[3mm] \sum_{n = 1}^{\infty}{1 \over 4n^{2} + 2n} &= {1 \over 4}\sum_{n = 1}^{\infty}{1 \over n\left(n + 1/2\right)} = {1 \over 4}\sum_{n = 0}^{\infty}{1 \over \left(n + 1\right)\left(n + 3/2\right)} = {1 \over 4}{\Psi\left(3/2\right) - \Psi\left(1\right) \over 3/2 - 1} \\[3mm]&= {1 \over 2} \left\lbrack\Psi\left(3 \over 2\right) - \Psi\left(1\right)\right\rbrack \\[5mm]& \end{align}

$$ \Psi\left(3 \over 2\right) = \underbrace{\Psi\left(1 \over 2\right)} _{-\gamma\ -\ 2\ln\left(2\right)} + {1 \over \left(1/2\right)} = -\gamma + 2\left\lbrack 1 - \ln\left(2\right)\right\rbrack\,, \qquad\qquad \Psi\left(1\right) = -\gamma $$

$$ \begin{array}{|c|}\hline\\ {\large\quad\sum_{n = 1}^{\infty}{1 \over 4n^{2} + 2n} = 1 - \ln\left(2\right)\quad} \\ \\ \hline \end{array} $$

$\endgroup$
  • $\begingroup$ How is $\Psi$ defined? $\endgroup$ – apnorton Sep 3 '13 at 22:00
  • $\begingroup$ @anorton $$ \Psi\left(z\right) \equiv {{\rm d}\ln\Gamma\left(z\right) \over {\rm d}z} $$ where $\Gamma\left(z\right)$ is the Gamma function. $\endgroup$ – Felix Marin Sep 3 '13 at 23:41
0
$\begingroup$

Hint: Another approach is to Complete the square in the denominator and use the Fourier transform and Poison summation formula approach.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.